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3 years ago

HELP ASAP WILL GIVE BRAINLIEST

Mathematics

1 answer:

____ [38]3 years ago

4 0

Answer:

1) x≠7

2) x≠3 or -7

Step-by-step explanation:

1. The given function is

$f(x) = \frac{1}{x - 7}$

This function is undefined if the denominator is equal to zero .

Therefore the restriction is that:

The denominator is not zero.

$x - 7 \ne0$

$x \ne7$

2) Assuming the second function is

$f(x) = \frac{(x + 7)(x - 9)}{(x - 3)(x + 7)}$

This function is not defined when the denominator is zero.

This implies that:

$(x - 3)(x + 7) \ne0$

The restrictions are:

$x \ne3 \: or \: - 7$

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find the formula for a rational function that satisfies all of the following. its x-intercept is (-3,0). it has a vertical asymp

Norma-Jean [14]

Let that be

$\\ \rm\Rightarrow y=\dfrac{a(f(x))}{g(x)}$

Two vertical asymptotes at -1 and 0

$\\ \rm\Rightarrow y=\dfrac{a(f(x)}{x(x+1)}$

If we simply

$\\ \rm\Rightarrow y=\dfrac{a(f(x)}{x^2+1}$

Denominator has degree 2
Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
But there should be a x intercept at -3 so one zeros should be -3

Find a equation

3x²+9x

Find zeros

3x²+9x=0
3x(x+3)=0
x=0,-3

Horizontal asymptote

3x²/x²
3

So our equation is

$\\ \rm\Rightarrow y=\dfrac{3x^2+9x}{x(x+1)}$

Graph attached

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1 year ago

Hello<br><img src="https://tex.z-dn.net/?f=.52.22.2." id="TexFormula1" title=".52.22.2." alt=".52.22.2." align="absmiddle" class

Rom4ik [11]

Uhh..not in a mood for a meet, k? Maybe later hun~~

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2 years ago

What is the area, in square meters, of the trapezoid below

ahrayia [7]

Answer:

114.12

Step-by-step explanation:

A = (a+b)/2 ×h

= (6.2 + 25.5)/2 ×7.2

=114.12

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2 years ago

Can someone answer this question please?

allochka39001 [22]

Answer:

Step-by-step explanation:9

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3 years ago

Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (naut

frozen [14]

Answer:

a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$

Notice that:

$4\sqrt{13}=\sqrt{12^2+5^2}$ =√(speed of ship A² + speed of ship B²)

5 0

3 years ago